University of California, Los Angeles Department of ...
University of California, Los Angeles Department of Statistics
Statistics 100B
Instructor: Nicolas Christou
Distributions related to the normal distribution
Three important distributions:
? Chi-square (2) distribution.
? t distribution.
? F distribution.
Before we discuss the 2, t, and F distributions here are few important things about the gamma () distribution. The gamma distribution is useful in modeling skewed distributions for variables that are not negative.
A random variable X is said to have a gamma distribution with parameters , if its probability density function is given by
x-1
e-
x
f (x) =
, , > 0, x 0.
()
E(X) = and 2 = 2.
A brief note on the gamma function: The quantity () is known as the gamma function and it is equal to:
() = x-1e-xdx.
0
Useful result: 1
( ) = . 2
If
we
set
=
1
and
=
1
we
get
f (x)
=
e-x.
We
see
that
the
exponential
distribution
is
a special case of the gamma distribution.
1
The gamma density for = 1, 2, 3, 4 and = 1. Gamma distribution density
1.0
0.8
0.6
( = 1, = 1)
( = 2, = 1) ( = 3, = 1)
( = 4, = 1)
f(x)
0.4
0.2
0.0
0
2
4
6
8
x
Moment generating function of the X (, ) random variable:
MX (t) = (1 - t)-
Proof: MX (t) = EetX =
etx
x-1e-
x
dx
=
1
0
()
()
x-1e-x(
1-t
)dx
0
Let
y
=
x(
1-t
)
x
=
1-t
y,
and
dx
=
1-t
dy.
Substitute
these
in
the
expression
above:
1
MX (t) = () 0
-1
y-1e-y
dy
1 - t
1 - t
1 MX(t) = ()
1 - t
-1 1 - t
y-1e-ydy MX (t) = (1 - t)-.
0
2
Theorem: Let Z N (0, 1). Then, if X = Z2, we say that X follows the chi-square distribution with 1 degree of freedom. We write, X 21.
Probability density function of X 21:
Find the probability density function of X
= Z2, where f (z) =
1
e-
1 2
z2
.
Begin with the
2
cdf of X:
FX (x)
=
P (X
x)
=
P (Z2
x)
=
P (- x
Z
x)
FX(x) = FZ( x) - FZ(- x). Therefore:
fX (x)
=
1
x-
1 2
2
1
e-
1 2
x
2
+
1
x-
1 2
2
1
e-
1 2
x
2
=
2
1 2
1
x-
1 2
e-
x 2
,
or
x e -
1 2
-
x 2
fX(x) =
2
1 2
(
1 2
)
.
This
is
the
pdf
of
(
1 2
,
2),
and
it
is
called
the
chi-square
distribution
with
1
degree
of
freedom.
We write, X 21.
The
moment
generating
function
of
X
21
is
MX (t)
=
(1
-
2t)-
1 2
.
Theorem:
Let Z1, Z2, . . . , Zn be independent random variables with Zi N (0, 1). If Y =
n i=1
zi2
then
Y follows the chi-square distribution with n degrees of freedom. We write Y 2n.
Proof: Find the moment generating function of Y . Since Z1, Z2, . . . , Zn are independent,
MY (t) = MZ12(t) ? MZ22(t) ? . . . MZn2 (t)
Each
Zi2
follows
21
and
therefore
it
has
mgf
equal
to
(1
-
2t)-
1 2
.
Conclusion:
MY
(t)
=
(1
-
2t)-
n 2
.
This
is
the
mgf
of
(
n 2
,
2),
and
it
is
called
the
chi-square
distribution
with
n
degrees
of
free-
dom.
Theorem: Let X1, X2, . . . , Xn independent random variables with Xi N (?, ). It follows directly form the previous theorem that if
n
Y=
i=1
xi - ? 2
then Y 2n.
3
f(x)
0.10 0.20
0.00
We know that the mean of (, ) is E(X) = and its variance var(X) = 2. Therefore, if X 2n it follows that:
E(X) = n, and var(x) = 2n. Theorem: Let X 2n and Y 2m. If X, Y are independent then
X + Y 2n+m. Proof: Use moment generating functions.
Shape of the chi-square distribution: In general it is skewed to the right but as the degrees of freedom increase it becomes N (n, 2n). Here is the graph:
32
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 x
120
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 x
320
0 4 8 12 16 20 24 28 32 36 40 44 48 52 56 60 64 68 72 76 80 84 88 92 96 x
4
f(x)
0.10 0.20
0.00
f(x)
0.10 0.20
0.00
The 2 distribution - examples Example 1 If X 216, find the following:
a. P (X < 28.85). b. P (X > 34.27). c. P (23.54 < X < 28.85). d. If P (X < b) = 0.10, find b. e. If P (X < c) = 0.950, find c. Example 2 If X 212, find constants a and b such that P (a < X < b) = 0.90 and P (X < a) = 0.05. Example 3 If X 230, find the following: a. P (13.79 < X < 16.79). b. Constants a and b such that P (a < X < b) = 0.95 and P (X < a) = 0.025. c. The mean and variance of X. Example 4 If the moment-generating function of X is MX(t) = (1 - 2t)-60, find: a. E(X). b. V ar(X). c. P (83.85 < X < 163.64).
5
Theorem: Let X1, X2, . . . , Xn independent random variables with Xi N (?, ). Define the sample variance as
S2
=
n
1 -
1
n
(xi
i=1
-
x?)2 .
Then
(n - 1)S2 2
2n-1.
Proof:
Example: Let X1, X2, . . . , X16 i.i.d. random variables from N (50, 10). Find
n
a.
P 796.2 < (Xi - 50)2 < 2630 .
i=1
n
b.
P 726.1 < (Xi - X? )2 < 2500 .
i=1
6
The t distribution
Definition: Let Z N (0, 1) and U 2df . If Z, U are independent then the ratio
Z follows the t (or Student's t) distribution with degrees of freedom equal to df .
U df
We write X tdf .
The probability density function of the t distribution with df = n degrees of freedom is
f (x)
=
(
n+1 2
n(
)
n 2
)
x2 1+
n
-
n+1 2
,
- < x < .
Let X tn.
Then,
E (X )
=
0
and
var(X )
=
n n-2
.
The t distribution is similar to the
standard normal distribution N (0, 1), but it has heavier tails. However as n the t
distribution converges to N (0, 1) (see graph below).
N(0, 1)
t15
t5
t1
f(x)
0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45
-5.0 -4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0 5.0 x
7
Application:
Let X1, X2, . . . , Xn be independent and identically distributed random variables each one
having
N (?, ).
We
have
seen
earlier
that
(n-1)S2 2
2n-1.
We
also
know
that
X? -?
n
N (0, 1).
We can apply the definition of the t distribution (see previous page) to get the following:
X? -?
n
(n-1)S2 2
n-1
X? - ? = s .
n
Therefore
X? -?
s
tn-1.
n
Compare
it
with
X? -?
N (0, 1).
n
Example: Let X? and SX2 denote the sample mean and sample variance of an independent random sample of size 10 from a normal distribution with mean ? = 0 and variance 2. Find c so
that
X?
P
< c = 0.95
9SX2
8
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